**Maths for Chemists**This page has a focus on mathematics in A-level chemistry. Below is a challenging question that came from a textbook in Vietnam.

__Question:__

Based on observations in the laboratory, a student writes the following unbalanced equation to describe the reduction of an unidentified metal oxide.

R_{x}O_{y} + CO --> R + CO_{2} *Equation 1*

The mass of the metal oxide is 8g and the volume of carbon monoxide that reacts is 3.36dm^{3} (at STP, ie: molar volume = 22.4 dm^{3} mol ^{-1})

All of the resulting metal is then reacted with hydrochloric acid in a second reaction, according to the following equation:

2R +2nHCl --> 2RCln + nH_{2} *Equation 2*

The volume of HCl reacted is 200cm^{3}, and the concentration is 1 mol dm^{-3}.

Identify the metal R.

** Solution**
:

*Abbreviations and data*
:

n = number of moles (AKA amount of substance)

V = volume / dm^{3}

M = mass / g

C = concentration / mol dm^{-3}

MM = molar mass / g mol^{-1}

22.4dm^{3} = molar volume (at STP)

*Solution:*

__1) Work out the total mass of the reactants__

Using the available data, we can work out the mass of carbon monoxide that reacted, which will enable us to calculate the total mass of reactants.

n (CO) = V / 22.4 = 3.36 / 22.4 = 0.15mol

m (CO) = n x MM = 0.15 x (12 + 16) = 4.2

total mass of products = total mass of reactants = 8 + 4.2 = 12.2g

__2) Work out the masses of the products__

Each side of the equation only has one carbon-containing compound (CO from the reactants and CO_{2} from the products), each of which contains just
one atom of carbon. Therefore, even though we do not know how the equation is balanced, the coefficients for the two carbon compounds must be the same, ie:
n (CO) = n (CO_{2}).

n (CO) = n (CO_{2}) = 0.15mol

m (CO_{2}) = n x MM = 0.15 x (12 + [2 x 16]) = 6.6g

Therefore the mass of the extracted metal produced can be determined by subtracting the mass of carbon dioxide from the total mass of the products.

m (R) = total mass products – m (CO_{2}) = 12.2 – 6.6 = 5.6g

**
Quick solution: Hunch: is the metal iron? Reasoning: the mass of R is a factor of 10 less than the molar mass of iron. Also, this equation is very
familiar because iron oxide is industrially reduced by carbon monoxide. Skip to step 4 to play the hunch.
**

__3) How does the metal react with hydrochloric acid in reaction 2? __

Remember that the volume of HCl reacted is 200cm^{3}, and the concentration is 1 mol dm^{-3}.

2R +2_{N}HCl --> 2RCl_{N} + _{N}H_{2} *Equation 2*

There may be a more sophisticated way to do this but trial and error is also effective.

n (HCl) = c x v = 1 x 200cm^{3} / 1000 = 0.2mol

From equation 2, N links the coefficients of HCl and H_{2} to the molecular formula of the metal chloride (RCl_{N}). N is not just a number
whose value can be revealed by applying mathematics, it also relates to chemical concepts. For example, if N were 20, equation 2 would be as follows:

2R + 40HCl --> 2RCl_{20} + 20H_{2}

In other words, we would be looking for a metal that can form bonds to 20 chlorine atoms, which does not seem very likely. A sensible upper limit to apply would be N = 6, which comes out as follows:

2R + 12HCl --> 2RCl_{6} + 6H_{2}

The transition metal tungsten can form complex ions with a formula of WCl_{6}, so, while unlikely, this equation is consistent with what can
reasonably be expected of tungsten. However, if we assign this value, there are implications for the relative mass of the mystery metal (R), as well as for
the mystery metal oxide (R_{x}O_{y}). In this case, if N = 6, the stoichiometry between R and HCl is 1:6. We know that 0.2 moles of HCl
reacted, so we can deduce the molar mass of R as follows:

n (R) = 2 / 12 x n (HCl) = 0.03333 (4sf)

MM (R) = m / n = 5.6 / 0.03333 = 168

This is very close to the relative mass of thulium, which is 168.9. Let’s suppose thulium is the mystery metal. Does it make thulium hexachloride? A web
search suggests no, the likely chloride is in fact TmCl_{3}. Also, the most typical oxide is Tm_{2}O_{3}, so let’s see if the sums
check out in equation 1.

Let R = Tm

Equation 1:

Tm_{2}O_{3} + 3CO --> 2Tm + 3CO_{2}

We know that the mass of the metal oxide is 8g and we know that the number of moles of CO was 0.15. The stoichiometry between the metal oxide and the CO is
1:3, meaning that the number of moles of Tm_{2}O_{3 }would be 0.05. So, would 0.05 moles of Tm_{2}O_{3} have a mass of 8g?
Let’s see.

MM Tm_{2}O_{3} = (168.9 x 2) + (3 x 16) = 385.8gmol^{-1}

Mass of 0.05 moles of Tm_{2}O_{3} = n x MM = 385.8 x 0.05 = 19.3g

The given mass and the calculated mass are different, from which we can infer two things: 1) the metal definitely is not thulium and 2) our metal must have a lower relative mass in order for the sums to check out.

Let’s try 1, 2 and 3 as potential values of N. We can get a spreadsheet to do the heavy lifting.

N |
coefficient HCl (2N) |
n (HCl) |
coefficient R |
n (R) |
MM R |

1 |
2 |
0.2 |
2 |
0.2 |
28 |

2 |
4 |
0.2 |
2 |
0.1 |
56 |

3 |
6 |
0.2 |
2 |
0.066667 |
84 |

These molar mass values could be matched to the following elements:

Silicon 28.1

Krypton, 83.80 / Rubidium 85.47

Iron 55.8

__4) Play the Hunch__

We could painstakingly go through each in turn, but it’s time to play a hunch. From the very beginning, this equation has been shouting ** Iron Oxide**. The reduction of iron oxide by carbon monoxide to produce iron and carbon dioxide will be familiar to chemistry
students. Meanwhile, the mass of R produced in equation 1 happens to be a convenient factor of 10 smaller than the molar mass of iron. (When the relative
mass of iron is expressed to zero decimal places.) Let’s play the hunch.

The chemical formula of iron (III) oxide is Fe_{2}O_{3}. As such, equation 1 can be written as follows:

Fe_{2}O_{3} + 3CO --> 2Fe + 3CO_{2}

As with thulium, we have a 1:3 stoichiometry between the metal oxide and the carbon monoxide. Recalling, n (CO) = 0.15, therefore n (R_{x}O _{y}) = 0.05

MM Fe_{2}O_{3} = (55.8 x 2) + (3 x 16) = 159.6gmol^{-1}

Mass of 0.05 moles of Fe_{2}O_{3} = n x MM = 0.5 x 159.6 = 7.98g

Since the question stated that the original mass of metal was 8g, we can assume this is the correct answer. Remember that the answer will be affected by
the number of decimal places adopted for the relative masses of each element. If we run the question with zero decimal place values, ie: relative mass Fe =
56gmol^{-1}, the final answer comes out at an exact 8g.

Finally, to be thorough, we can write out the completed equation 2:

2Fe + 4HCl --> 2FeCl_{2} + 2H_{2}

and then simplifying:

Fe + 2HCl --> FeCl_{2} + H_{2 }

This fits with the known formula for iron (II) chloride.

So, the metal R is iron.